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Find the MacLaurin Series for $f(x)=\ln(√81-x^2)$

I know that $\ln(\sqrt{81}-x^2) = 1/2(\ln(81-x^2)) = 1/2(\ln(9-x)) + 1/2(\ln(9+x))$

I need help finding the series, not the expansion.

Michael Hardy
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math123
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  • Note that $\ln(81-x^2)=\ln(81)+\ln\left(1-\frac{x^2}{81}\right)$. You probably know the Maclaurin series for $\ln(1+t)$. Replace $t$ everywhere by $-\frac{x^2}{81}$. – André Nicolas Apr 15 '15 at 04:43

2 Answers2

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Hint. You may just write $$ \ln (9-x)=\ln 9+\ln \left(1-\frac x9\right) $$ and use $$ -\ln (1-u)=u+\frac{u^2}2+\frac{u^3}3+\cdots+\frac{u^n}n+\cdots,\qquad |u|<1,\tag1 $$ by putting $u=\dfrac x9$.

You may do the same for $\ln (9+x)$.

Michael Hardy
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Olivier Oloa
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HINT:

$$\ln(9-x)=\ln9+\ln\left(1-\dfrac x9\right)$$

See https://math.stackexchange.com/questions/920095/series-expansion-of-log1x2

and Taylor series for $\log(1+x)$ and its convergence

Community
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lab bhattacharjee
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